Question

(1+3)(1+3^2)(1+3^4)(1+3^8)…(1+3^32)等于(1-3^64)/(-2)，是怎么计算的？请把计算过程详细点写出来，多谢

Answer 1

(1+3)(1+3^2)(1+3^4)(1+3^8)…(1+3^32)
=(1-3)(1+3)(1+3^2)(1+3^4)(1+3^8)…(1+3^32)/(1-3)
=(1-3²)(1+3^2)(1+3^4)(1+3^8)…(1+3^32)/(-2)
=(1-3^4)(1+3^4)(1+3^8)…(1+3^32)/(-2)
=(1-3^8)(1+3^8)…(1+3^32)/(-2)
=(1-3^16)(1+3^16)(1+3^32)/(-2)
=(1-3^32)(1+3^32)/(-2)
等于(1-3^64)/(-2)

Answer 2

解：(1+3)(1+3^2)(1+3^4)(1+3^8)…(1+3^32)
=-1/2×(1-3)(1+3)(1+3^2)(1+3^4)(1+3^8)…(1+3^32)
=-1/2×(1-3²)(1+3^2)(1+3^4)(1+3^8)…(1+3^32)
=-1/2×(1-3^4))(1+3^4)(1+3^8)…(1+3^32)
=-1/2×(1-3^16)(1+3^16)(1+3^32)
=-1/2×(1-3^32)(1+3^32)
=-1/2×(1-3^64)
=(1-3^64)/(-2)

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Answer 3

方程乘(1-3)再除(1-3)
(1+3)(1+3^2)(1+3^4)(1+3^8)…(1+3^32)

=(1+3)(1+3^2)(1+3^4)(1+3^8)…(1+3^32)*(1-3)/(1-3)
=(1+3)(1-3)(1+3^2)(1+3^4)(1+3^8)…(1+3^32)/(1-3)
=(1-3^2)(1+3^2)(1+3^4)(1+3^8)...(1+3^32)/(1-3)
=(1-3^4)(1+3^4)(1+3^8)...(1+3^32)/(1-3)
=(1-3^8)(1+3^8)...(1+3^32)/(1-3)
=(1-3^32)(1+3^32)/(-2)
=(1-3^64)/(-2)